Electrical Machines · 11 min read

DC machines

A DC machine converts mechanical energy to DC electrical energy (generator) or the reverse (motor). The same machine can do both. Although power electronics has replaced many DC drives, DC machines still appear in traction, cranes, battery-fed systems and as exciters, and their theory — emf equation, armature reaction, commutation, characteristics and speed control — is heavily tested in NEC exams.

Key concepts

  • Construction:
    • Yoke: outer frame, mechanical support and return path for flux (cast iron/steel).
    • Poles and pole shoes: carry field winding; shoe spreads flux and supports the coil.
    • Field winding: produces main flux.
    • Armature core: laminated silicon steel drum with slots; armature winding: lap (A=PA = P parallel paths, for high current/low voltage) or wave (A=2A = 2, for high voltage/low current).
    • Commutator: copper segments insulated by mica; acts as a mechanical rectifier (generator) or inverter (motor).
    • Brushes: carbon/graphite, collect current from the commutator.
    • Interpoles and compensating windings (in large machines).
  • Working principle: generator — conductors cut flux, emf by Fleming's right-hand rule, AC in the coil rectified by the commutator. Motor — current-carrying conductors in flux experience force (left-hand rule); the induced back emf EbE_b opposes supply voltage and regulates armature current.
  • Types by excitation: separately excited; self-excited — shunt, series, compound (cumulative/differential, long/short shunt).
  • Generator characteristics:
    • Shunt: voltage builds up only if residual magnetism exists, field connection aids it, and field resistance < critical resistance; speed above critical speed. Terminal voltage falls slightly with load.
    • Series: voltage rises with load (used as boosters).
    • Cumulative compound: flat (level) or over-compounded — used for supply over feeders.
  • Motor characteristics:
MotorSpeed–loadTorqueTypical use
ShuntNearly constant speedT∝IaT \propto I_aLathes, fans, pumps, conveyors
SeriesSpeed falls sharply; dangerous at no loadT∝Ia2T\propto I_a^2 (before saturation), high starting torqueTraction, cranes, hoists
Cumulative compoundBetween the two; safe no-load speedGood starting torquePresses, shears, rolling mills

A series motor must never be started or run without load (belt drive avoided).

  • Armature reaction: effect of armature MMF on main field. It cross-magnetises (distorts flux, shifts the magnetic neutral axis — forward in generator, backward in motor) and demagnetises (with brush shift). Remedies: compensating winding in pole faces, interpoles, high-reluctance pole tips.
  • Commutation: reversal of coil current while short-circuited by a brush; delayed reversal (reactance voltage) causes sparking. Remedies: resistance commutation (carbon brushes), emf commutation (interpoles, brush shift).
  • Losses: copper (armature, field, brush contact), iron (hysteresis + eddy in armature), mechanical (friction, windage). Constant losses = iron + mechanical + shunt field. Maximum efficiency when variable (armature Cu) loss = constant loss.
  • Starting: at standstill Eb=0E_b = 0, so Ia=V/RaI_a = V/R_a would be very large. A starter adds series resistance which is cut out as speed rises: 3-point starter (shunt motor; no-volt coil in field circuit — opens if field fails), 4-point starter (no-volt coil across supply, independent of field — allows field control). Both have overload release.
  • Speed control from N∝(V−IaRa)/ΦN \propto (V - I_aR_a)/\Phi:
    • Flux (field) control: rheostat in shunt field — above-rated speed, efficient, constant power.
    • Armature resistance control: below-rated speed, wasteful.
    • Armature voltage control: Ward-Leonard system or modern thyristor/chopper drives — smooth, wide range, both directions.
    • Series motor: diverter across field, tapped field, series–parallel control (traction).

High-yield points

  • Emf E=PΦZN/(60A)E = P\Phi ZN/(60A); lap winding A=PA = P, wave winding A=2A = 2.
  • Motor: V=Eb+IaRaV = E_b + I_aR_a; speed N∝Eb/ΦN \propto E_b/\Phi.
  • Series motor: T∝Ia2T \propto I_a^2 (unsaturated), highest starting torque, never run on no load.
  • 3-point starter's no-volt coil is in series with the field; 4-point starter's is across the supply.
  • Maximum efficiency when variable (armature copper) loss = constant loss.
  • Field weakening raises speed above rated; armature resistance control gives below-rated speed.

Formulas

E=PΦZN60AE = \frac{P\Phi Z N}{60A} P = poles, Φ = flux/pole (Wb), Z = total conductors, N = speed (rpm), A = parallel paths.

Generator: V=E−IaRaV = E - I_aR_a; motor: V=Eb+IaRaV = E_b + I_aR_a.

T=0.159PΦZIaA N⋅m,N∝EbΦT = 0.159\frac{P\Phi Z I_a}{A}\ \text{N·m},\qquad N \propto \frac{E_b}{\Phi}

Worked example

A 220 V shunt motor has Ra=0.5R_a = 0.5 Ω and draws armature current 20 A at 1000 rpm. Find speed when Ia=40I_a = 40 A (flux constant).

  1. Eb1=220−20×0.5=210E_{b1} = 220 - 20\times0.5 = 210 V.
  2. Eb2=220−40×0.5=200E_{b2} = 220 - 40\times0.5 = 200 V.
  3. N2=1000×200/210=952N_2 = 1000\times200/210 = 952 rpm.

Exam tips

  • Lap: A = P; wave: A = 2 regardless of poles.
  • Series motor: highest starting torque, never on no load. Shunt: constant speed.
  • Field weakening raises speed; armature resistance control lowers speed.
  • Back emf is what makes a DC motor self-regulating; starter needed because Eb=0E_b = 0 at start.
  • Interpoles are in series with the armature; compensating winding neutralises cross-magnetisation under pole faces.

Exam-style questions

From Calmon's question bank, which has 25 questions on this topic.

1. The armature of a DC motor has a back emf of 200 V and an armature current of 50 A while running at 1000 rpm. The electromagnetic torque developed is about:

  • A. 955 N·m
  • B. 191 N·m
  • C. 95.5 N·m
  • D. 10 N·m

Answer: C. T=EbIa/ω=(200×50)/(2π×1000/60)=10000/104.7≈95.5T = E_bI_a/\omega = (200\times50)/(2\pi\times1000/60) = 10000/104.7 \approx 95.5 N·m.

2. A 220 V DC motor has armature resistance 0.5 Ω0.5\,\Omega. The external resistance needed in the starter to limit the starting armature current to 40 A is:

  • A. 0.5 Ω
  • B. 5.5 Ω
  • C. 5.0 Ω
  • D. 4.5 Ω

Answer: C. At start Eb=0E_b = 0: total resistance =220/40=5.5 Ω= 220/40 = 5.5\,\Omega, so external resistance =5.5−0.5=5.0 Ω= 5.5 - 0.5 = 5.0\,\Omega.

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